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evablogger [386]

3 years ago

8

A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of L 0 m. If the initial vertical velocity

is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?

1 answer:

disa [49]3 years ago

3 0

Complete Question:

A volley ball is hit directly toward the ceiling in a gymnasium with a ceiling height of 10 m. If the initial vertical velocity is 13 m/s and the release height is 1.8 m will the ball hit the ceiling?

Answer:

The ball will hit the ceiling

Explanation:

Given;

Initial vertical Velocity U = 13 m/s

Height of the ceiling = 10 m

Released height of the volley ball = 1.8 m

Height traveled by the volley ball, is calculated as follows;

$V^2 =U^2 -2gH$

where;

V is final vertical velocity

$2gH =U^2\\\\H = \frac{U^2}{2g} = \frac{(13)^2}{2(9.8)} = 8.62 m$

Remember this ball was released from 1.8 m height and it traveled 8.62 m.

Total distance traveled = 1.8 + 8.62 = 10.42 m

Therefore, the ball will hit the ceiling

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Steel wire rope is used to lift a heavy object. We use a 3.1m steel wire that

kaheart [24]

Answer:

Young's modulus for the rope material is 20.8 MPa.

Explanation:

The Young's modulus is given by:

$E = \frac{FL_{0}}{A\Delta L}$

Where:

F: is the force applied on the wire

L₀: is the initial length of the wire = 3.1 m

A: is the cross-section area of the wire

ΔL: is the change in the length = 0.17 m

The cross-section area of the wire is given by the area of a circle:

$A = \pi r^{2} = \pi (\frac{0.006 m}{2})^{2} = 2.83 \cdot 10^{-5} m^{2}$

Now we need to find the force applied on the wire. Since the wire is lifting an object, the force is equal to the tension of the wire as follows:

$F = T_{w} = W_{o}$

Where:

$T_{w}$ : is the tension of the wire

$W_{o}$ : is the weigh of the object = mg

m: is the mass of the object = 1700 kg

g: is the acceleration due to gravity = 9.81 m/s²

$F = mg = 1700 kg*9.81 m/s^{2} = 16677 N$

Hence, the Young's modulus is:

$E = \frac{16677 N*0.006 m}{2.83 \cdot 10^{-5} m^{2}*0.17 m} = 20.8 MPa$

Therefore, Young's modulus for the rope material is 20.8 MPa.

I hope it helps you!

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Similarities in human fossils

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3. The load across a 12-V battery consists of a series combination

Evgesh-ka [11]

Answer:

a. 0.2 A

Explanation:

Given;

voltage of the battery, V = 12 V

three resistance connected in series, R₁ = 15 Ω, R₂ = 21 Ω, R₃ = 24Ω

The equivalent resistance for series connection is calculated as;

Rt = R₁ + R₂ + R₃

Rt = 15Ω + 21Ω + 24Ω

Rt = 60 Ω

Apply Ohm's law to calculate the current n the circuit;

V = IRt

I = V/Rt

I = 12/60

I = 0.2 A

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Speedy Sue, que conduce a 30.0 m/s, entra a un túnel de un carril. En seguida observa una camioneta lenta 155 m adelante que se

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Answer:

Si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

Explanation:

Supongamos que el vehículo de Speedy Sue decelera a razón constante, mientras que la camioneta se desplaza a velocidad constante. Se requiere conocer si ambos vehículos colisionarán, lo cual implica conocer si existe algún instante tal que ambos tengan la misma posición. Consideremos además que la posición de referencia se encuentra en la posición inicial de Sppedy Sue. Entonces, las ecuaciones cinemáticas son:

Speedy Sue

$x_{S} = x_{S,o}+v_{S,o}\cdot t+\frac{1}{2}\cdot a_{S}\cdot t^{2}$

Camioneta lenta

$x_{C} = x_{C,o} +v_{C}\cdot t$

Donde:

$x_{S,o}$ , $x_{C,o}$ - Posiciones iniciales de Speedy Sue y la camioneta lenta, medidas en metros.

$v_{S,o}$ - Velocidad inicial de Speedy Sue, medida en metros por segundo.

$v_{S}, v_{C}$ - Velocidades actuales de Speedy Sue y la camioneta lenta, medidas en metros por segundo.

$a_{S}$ - Deceleración de Speedy Sue, medida en metros por segundo cuadrado.

$t$ - Tiempo, medido en segundos.

Si conocemos que $x_{S} = x_{C}$ , $x_{S,o} = 0\,m$ , $x_{C,o} = 155\,m$ , $v_{S,o} = 30\,\frac{m}{s}$ , $v_{C} =-5\,\frac{m}{s}$ y $a_{S} = -2\,\frac{m}{s^{2}}$ , encontramos la siguiente función cuadrática:

$155\,m + \left(-5\,\frac{m}{s} \right)\cdot t = 0\,m+\left(30\,\frac{m}{s} \right)\cdot t +\frac{1}{2}\cdot (-2\,\frac{m}{s^{2}} )\cdot t^{2}$

$-t^{2}+35\cdot t-155 = 0$ (Ec. 1)

Las raíces de esta función son:

$t_{1}\approx 29.798\,s$ , $t_{2} \approx 5.201\,s$

La colisión ocurriría en la raíz positiva más pequeña, es decir:

$t \approx 5.201\,s$

Ahora, la posición en que ocurriría la colisión se determina a partir de la ecuación de desplazamiento de la camioneta lenta, es decir: ( $v_{C,o} = -5\,\frac{m}{s}$ , $x_{C,o} = 155\,m$ , $t \approx 5.201\,s$ )

$x_{C} = 155\,m +\left(-5\,\frac{m}{s}\right)\cdot (5.201\,s)$

$x_{C} = 128.995\,m$

En síntesis, si ocurre una colisión. 5.201 segundos después de que Speedy Sue ingrese al túnel y a una distancia de 128.995 metros.

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